sir,how could this be solved,
there are 10 pairs of socks,in a cupboard from which 4 individual socks are picked at random.the probablity that there is at least one pair is.
The sample space would be 20C4. The number of ways of picking 4 socks with no matching pair would be 20 x 18 x 16 x 14 / 4!. Find the probability accordingly. Then, the probability of at least one pair is 1 – the above.
It can be looked at as the same logic as integration – area under the curve. At 5:00 the probability is 1/6, at 5:10 it is 1/3 and in between it increases linearly. So the average probability at any time in that interval will be just the average of the two extremes. Had it not been a linear increase then it would have been a different (and more complicated) story.
Hi J, this is a beautiful perspective you have added to my understanding of this subject; sincerely thanking you for taking the initiative to impart out-of-the-box education 🙂
sir,how could this be solved,
there are 10 pairs of socks,in a cupboard from which 4 individual socks are picked at random.the probablity that there is at least one pair is.
The sample space would be 20C4. The number of ways of picking 4 socks with no matching pair would be 20 x 18 x 16 x 14 / 4!. Find the probability accordingly. Then, the probability of at least one pair is 1 – the above.
regards
J
J, why did you divide by 4!?
Why do we take average of both the probabilities to calculate probability of first and the last interval?
It can be looked at as the same logic as integration – area under the curve. At 5:00 the probability is 1/6, at 5:10 it is 1/3 and in between it increases linearly. So the average probability at any time in that interval will be just the average of the two extremes. Had it not been a linear increase then it would have been a different (and more complicated) story.
regards
J
Got it..thanks!!
Hi J, this is a beautiful perspective you have added to my understanding of this subject; sincerely thanking you for taking the initiative to impart out-of-the-box education 🙂